$X$ and $Y$ have the same cardinality if and only if there exist a bijection from $X$ to $Y$? [duplicate]

My textbook says "Let $$X$$ and $$Y$$ be sets. We say $$X$$ and $$Y$$ have the same cardinality if there is a bijection $$f: X \to Y$$."

I was wondering why the text does not say "if and only if." A bijection implies same cardinality, but does cardinality imply bijection? I would imagine so.

marked as duplicate by Najib Idrissi, user642796Nov 1 '14 at 15:54

• Ignorant I read: "the sets $X$ and $Y$ have the same cardinality..." Hmm, what is meant by that? Let's have a look at the definition then. Oh.., I see. It means that there is a bijection between them. – drhab Nov 1 '14 at 15:36
it depends on your definition of cardinality, you can take that as a definition, or if you define $|A|\leq|B|$ iff the is an injective function $A\rightarrow B$, then it is a deep theorem that $|A|\leq|B|$ and $|B|\leq|A|$ then there is a bijective function between them, so "up to bijection" this defines an order.
A very nice example is a class with 10 seats and 10 students. Both sets have 10 elements and you can establish a bijection of giving a student exactly one seat. This is the intuitive idea of 2 sets of the same cardinality implying a bijection between the sets. If you can assign through $f:A\rightarrow B$ to every element $x$ in $A$ an element $y$ in $B$, and to every element $y$ in $B$ back to an element $x$ in $A$, then both sets have the same cardinality: there is a correspondence between elements of the sets which is a bijection.
For instance consider $g:\mathbb N \rightarrow \mathbb O$ being the last set the set of positive odd numbers then $g(n)=2n+1$ is a bijection. Therefore, even though $\mathbb O\subsetneq \mathbb N$, $|\mathbb O | = |\mathbb N|$.